Here is what I've come up with:
First I convert the date to a SAS date, which is the number of days since Jan. 1 1960:
data ds;
set ds (rename=(date=old_date));
date = input(old_date, date11.);
drop old_date;
run;
Then compute lag_sale (I am using the same calculation you used in the question, but make sure this is what you want to do. For some observations the lag sale is the previous recorded date, but for some it is the same store_id and date, just a different observation.):
proc sort data=ds; by store_id; run;
data ds;
set ds;
by store_id;
lag_sale = lag(sales);
if first.store_id then lag_sale = .;
run;
Then set up the final data set:
data final;
length store_id 8 date 8 cov 8 corr 8;
if _n_ = 0;
run;
Then create a macro which takes a store_id and date and runs proc corr. The first part of the macro selects only the data with that store_id and within the past 45 days of the date. Then it runs proc corr. Then it formats proc corr how you want it and appends the results to the "final" data set.
%macro corr(store_id, date);
data ds2;
set ds;
where store_id = &store_id and %eval(&date-45) <= date <=&date
and lag_sale ne .;
run;
proc corr noprint data=ds2 cov outp=corr;
by store_id;
var sales lag_sale;
run;
data corr2;
set corr;
where _type_ in ('CORR', 'COV') and _name_ = 'sales';
retain cov;
date = &date;
if _type_ = 'COV' then cov = lag_sale;
else do;
corr = lag_sale;
output;
end;
keep store_id date corr cov;
run;
proc append base=final data=corr2 force; run;
%mend corr;
Finally run the macro for each store_id/date combination.
proc sort data=ds out=ds3 nodupkey;
by store_id date;
run;
data _null_;
set ds3;
call execute('%corr('||store_id||','||date||');');
run;
proc sort data=final;
by store_id date;
run;